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This is a question from the book "Statistical Methods (6e) by Snedecor, Cochran":

10 balls, numbered 0 to 9, are placed in a box. For our sample, a ball is drawn from the box and then replaced in it. Our sample is built by drawing a ball from the box 10 times and noting the no. of odd numbers. This experiment is repeated 200 times.

Here is the frequency distribution obtained : enter image description here

The theoretical frequency has been obtained using the Equation of binomial distribution: enter image description here

For ex:
Theoretical freq for Class 0 = 200 x P(X=0 | 10,0.5) = 200 x 0.000977 ≈ 0.02
Theoretical freq for Class 1 = 200 x P(X=1 | 10,0.5) = 200 x 0.009766 ≈ 2

Here's the question from the book:
For the 200 samples of size 10 in the freq table, in how many cases is
a) the 95% confidence interval statement wrong?
b) the 99% confidence interval statement wrong?

The following table is given for 95% Confidence Interval (%) for Binomial Distribution: enter image description here

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  • $\begingroup$ Is this any different of a problem than flipping a coin ten times and counting the number of heads? $\endgroup$ – Dave Sep 26 at 13:27
  • $\begingroup$ No, it is similar. $\endgroup$ – gyaan.anveyshak Sep 26 at 13:29
  • $\begingroup$ Unclear exactly what you're asking. Is the first column $n=10$ and level 95% in the posted table the only thing of concern? Does interval for $f=0$ given as 0 27 mean $(0, 0.27)?$ // What style of binomial CI are you using: Wald? Agresti? Wilson? Jeffreys? ... What do you mean by "wrong"? Interval doesn't contain success probability? Whatever formula applies is incorrectly computed? Is there any connection btw 1st and 2nd tables in your Q? // Could you show what you have tried for even one part of the problem so we can see what you're trying and why you are asking for help? // Please edit Q. $\endgroup$ – BruceET Sep 27 at 2:33
  • $\begingroup$ @BruceET The question is being asked in the book. Not by me. I also difficulty understanding exactly what was being asked. Hence I posted it here thinking that I was missing something. $\endgroup$ – gyaan.anveyshak Sep 27 at 22:41
  • $\begingroup$ I have been able to replicate your first table using R. Now ready to figure out what the conf interval question is asking. What formula does your book use for CIs? What is the general topic under discussion? (Chapter heading, section heading, example heading, etc.) Is the word wrong theirs or yours; if theirs, look for nearby usages for a wrong CI and quote the sentence; if yours, please explain what you mean. [I feel like I've been put down in the middle of a game without knowing the rules or which side I'm playing for.] // I have S&C, 7e; there's nothing like this in Sec 1.4. $\endgroup$ – BruceET Sep 28 at 0:46
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First Table of your Question.

After some additional thought (following my Comment) I think I may have made sense of this.

First, consider the experiment with $m = 200$ samples (with replacement) of $n = 10$ balls from the box. Each sample give $X$ odd numbered balls, where $X \sim \mathsf{Binom}(n = 10, p = 1/2).$ Possible values of $X$ are integers $0$ through $10.$

Simulating $m = 200$ such experiments, I get the results tabled below:

set.seed(927)  # for reporducibility
x = rbinom(200, 10, .5)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  1.000   4.000   5.000   4.885   6.000   9.000 
table(x)
x
 1  2  3  4  5  6  7  8  9   # values 
 1 16 27 34 53 34 24  9  2   # frequencies

When Snedecor and Cochran did their parallel experiment they got the frequencies $1,1,8,25,39, \dots, 4, 0$ given in the middle column of the first table of your Question. The main theme of this section of the book seems to be sample variability. Their sample and mine are not the same, but were generated by equivalent procedures. Of course, both sets of frequencies sum to $m = 200.$

As you say, the expected frequencies for each value (or 'in each class') is obtained by multiplying the PDF of $\mathsf{Binom}(10, 1/2)$ by $m = 200.$ So the third column in the first table can be obtained in R as follows (ignore row numbers in [ ]s):

k = 0:10;  exp = 200*dbinom(k, 10, .5)
cbind(k,round(pdf,4),round(exp,2))
       k            
 [1,]  0 0.001  0.20
 [2,]  1 0.010  1.95
 [3,]  2 0.044  8.79
 [4,]  3 0.117 23.44
 [5,]  4 0.205 41.02
 [6,]  5 0.246 49.22
 [7,]  6 0.205 41.02
 [8,]  7 0.117 23.44
 [9,]  8 0.044  8.79
[10,]  9 0.010  1.95
[11,] 10 0.001  0.20

Table 1.4.1 from S&C.

In sampling $n = 10$ balls from the urn, you can get any one of 11 values: $X = 0, 2, 2, \dots, 10.$ Consequently, if you use such results to make a 95% CI for the binomial success probability $p,$ you will have any one of $11$ different confidence intervals, depending on the outcome of the experiment. Some of these 95% CIs will include the value $p = 1/2$ and some will not. We can call the ones tha cover 'good' and the others 'bad'

I don't know what style of CI S&C are using, but I can show results from one reasonably accurate style of CI. (This is done with the caveat that $n = 10$ is a very small sample, so we can anticipate some 'wide' confidence intervals.)

A 95% Jeffreys CI is based on a Bayesian argument, but it is often used by frequentist statisticians because it has good properties. If there are $x$ Successes in $n$ trials, this CI is found by taking quantiles $0.025$ and $0.975$ of the distribution $\mathsf{Beta}(x+.5, n-x+.5).$

So lower confidence limits lcl and upper confidence limits ucl of such a CI can be found in R as below. Table 1.4.1 expresses these limits in percentages, so I multiply by $100.$

k = 0:10;  n = 10
lcl = round(qbeta(.025, k+.5, n-k+.5),3)
ucl = round(qbeta(.975, k+.5, n-k+.5),3)
LCL = 100*lcl;  UCL = 100*ucl
cbind(k, LCL, UCL)
       k  LCL   UCL
 [1,]  0  0.0  21.7
 [2,]  1  1.1  38.1
 [3,]  2  4.4  50.3
 [4,]  3  9.3  60.6
 [5,]  4 15.3  69.6
 [6,]  5 22.4  77.6
 [7,]  6 30.4  84.7
 [8,]  7 39.4  90.7
 [9,]  8 49.7  95.6
[10,]  9 61.9  98.9
[11,] 10 78.3 100.0

Lower limits $0.0, 1.1, 4.4, 9.3, \dots, 78.3$ are roughly the same as the lower limits in the column n = 10 of Table 1.4.4, which are $(0, 0, 3, 8, \dots, 73.$ UCLs are also comparable.

Then, the issue arises, which ones of the 11 possible CIs include probability $1/2$ or $50$ percent (i.e, are 'good'). For my Jeffreys CIs, the answer is the CIs for 3 through 8 successes. [For the CIs given in S&C's table, the 'good' ones are the same.]

Finally, the 'coverage probability' of a CI for a particular probability (1/2 here) is the sum of the binomial probabilities that match 'good' CIs. That's about 98%. Because of the discreteness of the possible results, it turns out that there are 'lucky' and 'unlucky' values of the success probability $p$. One hopes that for most values of $p$ the coverage probability is near the promised '95%'.

In this respect, especially for small $n,$ 95% Jeffreys-style of CI has reasonably close to 95% coverage probability for almost all values of $p$ with $.25 < p < .75.$ [The Wald CI, which is of the form $\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}},$ where $\hat p = x/n,$ is intended for use with large $n,$ so it often performs poorly for small $n.]$

sum(dbinom(2:8, 10, .5))
[1] 0.9785156

Below is a plot of the coverage probabilities for the a 95% Jeffreys interval for $n = 10$ observations. The plot uses 2000 Success probabilities $p.$ Above, we have found the coverage probability $0.979$ at $p=1/2.$ (For $n = 10,$ almost all Wald coverage probabilities are below 95%, some far below; graph not shown.)

enter image description here

Ref: You can look at Wikipedia on binomial confidence intervals, including Jeffreys, Wald, and several others. Also relevant: Brown, et al. (2001): Interval estimates for a binomial proprotion, 16, Nr 2, p101-133, Statistical Science.


R code for figure:

n = 10; alp = .05; k = qnorm(1-alpha/2); x = 0:n
lcl = qbeta(alp/2,x+.5,n-x+.5)
ucl = qbeta(1-alp/2,x+.5,n-x+.5)

m = 2000;  pp=seq(1/n, 1 - 1/n, length=m)
p.cov = numeric(m)
for(i in 1:m) {
 cov = (pp[i] >= lcl) & (pp[i] <= ucl)
 p.rel = dbinom(x[cov], n, pp[i]) 
 p.cov[i] = sum(p.rel) }
plot(pp, p.cov, type="l", ylim=c(1-4*alp, 1),
     ylab="Coverage Probability", xlab="Success Probability",
     main="Jeffreys Interval Estimates: n = 10")
 abline(h = 1-alp, col="green2")

enter image description here

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